Reputation: 2977
With data frame:
df <- data.frame(id = rep(1:3, each = 5)
, hour = rep(1:5, 3)
, value = sample(1:15))
I want to add a cumulative sum column that matches the id
:
df
id hour value csum
1 1 1 7 7
2 1 2 9 16
3 1 3 15 31
4 1 4 11 42
5 1 5 14 56
6 2 1 10 10
7 2 2 2 12
8 2 3 5 17
9 2 4 6 23
10 2 5 4 27
11 3 1 1 1
12 3 2 13 14
13 3 3 8 22
14 3 4 3 25
15 3 5 12 37
How can I do this efficiently? Thanks!
Upvotes: 66
Views: 65913
Reputation: 4425
Using base R
df <- data.frame(id = rep(1:3, each = 5)
, hour = rep(1:5, 3)
, value = sample(1:15))
transform(df , csum = ave(value , id , FUN = cumsum))
#> id hour value csum
#> 1 1 1 4 4
#> 2 1 2 12 16
#> 3 1 3 13 29
#> 4 1 4 6 35
#> 5 1 5 5 40
#> 6 2 1 15 15
#> 7 2 2 1 16
#> 8 2 3 2 18
#> 9 2 4 8 26
#> 10 2 5 9 35
#> 11 3 1 11 11
#> 12 3 2 7 18
#> 13 3 3 10 28
#> 14 3 4 3 31
#> 15 3 5 14 45
Created on 2022-06-05 by the reprex package (v2.0.1)
Upvotes: 3
Reputation: 263352
df$csum <- ave(df$value, df$id, FUN=cumsum)
ave
is the "go-to" function if you want a by-group vector of equal length to an existing vector and it can be computed from those sub vectors alone. If you need by-group processing based on multiple "parallel" values, the base strategy is do.call(rbind, by(dfrm, grp, FUN))
.
Upvotes: 69
Reputation: 23757
Using dplyr::
require(dplyr)
df %>% group_by(id) %>% mutate(csum = cumsum(value))
Upvotes: 29
Reputation: 193527
To add to the alternatives, data.table
's syntax is nice:
library(data.table)
DT <- data.table(df, key = "id")
DT[, csum := cumsum(value), by = key(DT)]
Or, more compactly:
library(data.table)
setDT(df)[, csum := cumsum(value), id][]
The above will:
data.frame
to a data.table
by reference[]
there) the result of the entire operation"df" will now be a data.table
with a "csum" column.
Upvotes: 31
Reputation: 98449
Using library plyr
.
library(plyr)
ddply(df,.(id),transform,csum=cumsum(value))
Upvotes: 8